Let the amount invested in the two accounts be Rs. x and Rs. y respectively
By the given data, x + y = 8600   ..(1)
\(4\cfrac { 3 }{ 4 } \times \cfrac { x }{ 100 } +6\cfrac { 1 }{ 2 } \times \cfrac { y }{ 100 } =431.25\) \(\left[ \therefore interest=\cfrac { PNR }{ 100 } \right] \)
\(\Rightarrow \cfrac { 19x }{ 400 } +\cfrac { 13y }{ 3200 } =431.25\)
\(\Rightarrow \cfrac { 19x+26y }{ 400 } =431.25\)
19x + 26y = 172500   ...(2)
\(\Delta =\left| \begin{matrix} 1 & 1 \\ 19 & 26 \end{matrix} \right| =1(26)-1(19)\)
= 26-19 =7
\({ \Delta }x=\left| \begin{matrix} 8600 & 1 \\ 172500 & 26 \end{matrix} \right| =8600(26)-1(172500)\)
= 223600 - 172500 = 51100
\(\Delta y=\left| \begin{matrix} 1 & 8600 \\ 19 & 172500 \end{matrix} \right| =1(172500)-19(8600)\)
= 172500 - 163400 = 9100
\(x=\cfrac { \Delta x }{ \Delta } -\cfrac { 51100 }{ 7 } =7300\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 9100 }{ 7 } =1300\)
\(\therefore\) Investment in the interest of \(4\frac { 3 }{ 4 } \) % account is Rs. 7300 and investment in the rate of \(6\frac { 1 }{ 2 } \) account is Rs.1300. 

Let ‘x’ be the cost of a pencil
Let ‘y’ be the cost of an eraser
\(\therefore \) By given data, we get the following equations
11x + 3y = 64
8x + 3y = 49
\(\triangle =\left| \begin{matrix} 11 & 3 \\ 8 & 3 \end{matrix} \right| =9\neq 0,\) It has unique solution.
\({ \triangle }_{ x }\left| \begin{matrix} 64 & 3 \\ 49 & 3 \end{matrix} \right| =45\)
\({ \triangle }_{ y }\left| \begin{matrix} 11 & 64 \\ 8 & 49 \end{matrix} \right| =27\)
\(\therefore \) By Cramer’s rule
\(x={ \frac { \triangle x }{ \triangle } =\frac { 45 }{ 9 } =5 }\)
\(y={ \frac { \triangle y }{ \triangle } =\frac { 27 }{ 9 } =3 }\)
\(\therefore \) The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Let the price of one share of A be x
Let the price of one share of B be y
\(\therefore \) By given data, we get the following equations
10x + 5y = 125
9x + 12y = 150
\(\triangle =\left| \begin{matrix} 10 & 5 \\ 9 & 12 \end{matrix} \right| =75\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 125 & 5 \\ 150 & 12 \end{matrix} \right| =750\)
\({ \triangle }_{ y }=\left| \begin{matrix} 10 & 125 \\ 9 & 150 \end{matrix} \right| =375\)
\(\therefore \) Cramer’s rule 
\(x=\frac { \triangle x }{ \triangle } =\frac { 750 }{ 75 } =10\)
\( y=\frac { \triangle y }{ \triangle } =\frac { 375 }{ 75 } =5\)
The price of the share A is Rs10 and the price of the share B is Rs. 5.

The equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

Given 
A = \(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B = \(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
\(AB=\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-2+5 & -2+4-1 & 3-6+1 \\ 2+6+20 & -4-12+45 & 6+18-4 \\ 3+4+15 & -6-8+3 & 9+12-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) =\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (AB)=2\)
Now  \(BA=\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-4+9 & 1+6-6 & -1-8+9 \\ -2+8-18 & -2-12+12 & 2+16-18 \\ 5+2-3 & 5-3+2 & -5+4-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)

The number of non-zero rows is 2.
\(\therefore \rho (BA)=2\)